Joy uses 20n of force to shovel the snow 10 meters. how much work does she do? 200 j 2 j 30 j 100 j
1 answer:
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Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.
By Newton's second law, the net vertical force is
• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0
where a is the acceleration of the wagon.
Solve for n (the magnitude of the normal force) :
n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N
Then
f = 0.500 (58.0 N) = 29.0 N
Meanwhile, the horizontal component of the applied force has magnitude
(80.0 N) cos(30.0°) ≈ 69.3 N
Now calculate the work done by either force.
• friction: -(29.0 N) (10.0 m) = -290. J
• pull: (69.3 N) (10.0 m) = 693 J
Answer:
Explanation:
Given
coefficient of kinetic friction=0.34
inclination
weight of block=51 N
(a) When block is moving upward friction force acts downward
thus
as block is moving with constant velocity thus is zero
(b)When Block slides down the wall friction changes its direction to oppose the block